Q on associative binary operation
hi, Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}. *|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S. So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition? Regards Sarath. __________________________________ Do you Yahoo!? Yahoo! SiteBuilder - Free, easy-to-use web site design software http://sitebuilder.yahoo.com
On Thursday, August 28, 2003, at 12:14 AM, Sarad AV wrote:
hi,
Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}.
*|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d
The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S.
So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?
This looks like a homework assignment for a class. If a, b, c, and d are all arbitrary symbols, most substitutions (such as a = 2, b = 5, etc.) certainly will _not_ give (a*d)*d=a*(d*d)=d as a true statement. Only special values of a and d will make that true. (For example, a = 1 and d = 1 makes (1*1)*1=1*(1*1)=1. Other values may as well. Finding them is up to you. Lastly, your English is again unclear. "So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?" isn't a proper English sentence. Why do you keep posing these problems to the list? Are they homework problems? Do you think the list is just too quiet and needs ill=phrased puzzlers to keep it occupied? Did you ever do the coin flip experiment we suggested you do? Are you an AI which has failed the Turing Test and escaped? --Tim May
On Thu, Aug 28, 2003 at 12:14:20AM -0700, Sarad AV wrote:
hi,
Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}.
*|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d
The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S.
So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?
a d d d \ / \ / d d a d \ / \ / d = d What's the problem?
hi, Let ~ represents a relation. If a~b and b~a,then a~a (by transitivity) is an incorrect argument. By definition of transitivity, if a~b and b~c implies that a~c. I was asking on the same lines if (a*d)*d=a*(d*d)=d. By definition associativity is defined on a,b,c element of set S and not two elements of the set. x*y (ie, left*top) can be followed. Regards Sarath. --- BillyGOTO <billy@dadadada.net> wrote:
On Thu, Aug 28, 2003 at 12:14:20AM -0700, Sarad AV wrote:
hi,
Table shown is completed to define 'associative' binary operation * on S={a,b,c,d}.
*|a|b|c|d --------- a|a|b|c|d --------- b|b|a|c|d --------- c|c|d|c|d --------- d|d|c|c|d
The operation * is associative iff (a*b)*c=a*(b*c) for all a,b,c element of set S.
So can (a*d)*d=a*(d*d)=d considered as associative over * for this case as per definition?
a d d d \ / \ / d d a d \ / \ / d = d
What's the problem?
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On Thu, Aug 28, 2003 at 11:21:08PM -0700, Sarad AV wrote:
hi,
Let ~ represents a relation.
If a~b and b~a,then
a~a (by transitivity) is an incorrect argument.
By definition of transitivity, if a~b and b~c implies that a~c.
right.
I was asking on the same lines if (a*d)*d=a*(d*d)=d.
What does that have to do with transitivity? You didn't mention transitivity when you posed the question. Ridiculous. '*' is an operator, not a relation. Relations can't be parenthesized unless you're going to make truth or falsehood a symbol to be operated upon. Tim is right. Cypherpunks isn't a place to look for help with your algebra homework. I like doing interesting math problems, but you're not even properly asking the questions you want answered. That makes it a LOT less fun.
By definition associativity is defined on a,b,c element of set S and not two elements of the set.
This is getting stupid. The '*' operator was defined associative. The property of associativity applies to ASSOCIATIONS between symbols (i.e binary operators).
x*y (ie, left*top) can be followed.
I'm totally done with this.
participants (3)
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BillyGOTO -
Sarad AV -
Tim May